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Question

If a pair of variable straight lines x2+4y2+αxy=0 (where α is a real parameter) cut the ellipse x2+4y2=4 at two points A and B, then the locus of the point of intersection of tangents at A and B is


A

(2x-y)(2x+y)=0

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B

x2-4y2+8xy=0

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C

x2-4y2+4xy=0

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D

(x-2y)(x+2y)=0

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Solution

The correct option is D

(x-2y)(x+2y)=0


Step 1: Write equation for AB

Let the locus for intersection of tangents at A,B be P(h,k)

Equation of AB is given as xha+ykb=1

Here a=4,b=1

AB:xh4+yk1=1 ...(i)

Step 2: Homogenise the ellipse

x24+y21=xh4+yk12

x2h2-416+y2k2-1+hkxy2=0 ...(ii)

Given pair of straight lines is x2+4y2+αxy=0 ...(iii)

Step 3 : Solve for h,k

Equations (ii),(iii) represent the same lines.

Comparing the coefficients we get

h2-416=k2-14=hk2α

h2-4=4(k2-1)

h2-4k2=0

h+2kh-2k=0

Replace h,k with x,y to obtain the locus

x+2yx-2y=0 is the locus of point

Hence, option (D) is the correct answer


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